REVIEWER
FOR
CHEMICAL ENGINEERING
LICENSURE
EXAMINATION
3rd Edition
Solutions Manual
Monroe H. de Guzman, BSChE
Batangas State University
If some of the values are different, then these are brought by rounding off.
As much as possible, values are not rounded-off to obtain the most
possible accurate answers.
,
,I. Physical and Chemical Principles
A. General Inorganic Chemistry
4. For Zn2+, atomic number = 30, atomic mass = 65.38 (~65),
Number of protons = 30; neutrons = 65-30 = 35; electrons = 30 – 2 = 28
e- + proton + neutron = 30 + 35 + 28 = 93
6. The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six
different named series describing the spectral line emissions of the hydrogen atom (Wikipedia, 2014).
𝑛2
𝜆=( )(𝑅∞ /4); for m = 2 as n ∞
𝑛2−𝑚2
∞
Balmer series = 1/𝜆 = (𝑅∞ /4) = 10973731.57 m/4 = 27434.3289 cm -1
∞
7. Named after the American physicist Frederick Sumner Brackett who first observed the spectral
lines in 1922 (Wikipedia, 2014).
1 1
Brackett series = ( 2 − 2)(𝑅∞ ); m = 4 as n ∞ = 10973731.57 m/16 = 6858.5822 cm -1
𝑚 𝑛
10. Minimum wavelength of light for work functions (de Broglie wavelength)
For Φ = 2.90 eV (J/C of electron)
ℎ𝑐𝑜 (6.6261 𝑥 10−34 )(299792458)
𝜆= = = 427.5317 nm
𝜙𝑒 (2.90)(1.6022 𝑥 10−19 )
11. For Φ = 5.00 eV
ℎ𝑐𝑜 (6.6261 𝑥 10−34 )(299792458)
𝜆= = = 247.9584 nm
𝜙𝑒 (5.00)(1.6022 𝑥 10−19 )
13. Work function and kinetic energy
Rb Φ = 2.16 eV; λ = 350 nm of light wavelength.
Ephoton = Φ + Ek = hv = hco/λ; Ek = hco/λ – Φ
(6.6261 𝑥 10−34 )(299792458)
EK = − 2.16(1.6022 𝑥 10−19 )
350 𝑋 10−9
EK = 2.1667 x 10-19 J
Ek = (mev2)/2 = 689721.8665 m/s ~ 7.00 x 105 m/s
14. for n =2 and ms = -1/2. For l=0 and l = 1, it is composed of 3 counter-clockwise at p orbital and
1 at the s-orbital = 4 electrons
15. Cl = 35.45 composed of Cl-35 and Cl-37
35.45 = 37(x) + 35(1-x) where x is the fraction of Cl-37
x= 0.225; Cl-35 is 77.5 % in abundance.
17. If l = 0, ml should be = 0.
18. Outermost shell is at n = 4; taking the electrons at the outermost shell of 4s 2 and 4p3, 2 + 3=5.
31. T = 1600 °C; Br2(g) 2Br(g)
nBr2 = 1.05 moles ; Tv = 2 L; α = 0.025 Br2(g) 2Br(g)
[Br2] = 1.05/2 = 0.525 M
0.525 0
Kc = {([Br]2)/Br
2}eq = 1.3462 x M 10-3 0.525(1-α) 0.525x2α
Kp = Kc(RT)Σv
Kp = 1.3462 x 10-3(0.08206 x 1873.15)2-1 = 0.2069 atm 0.511875 M 0.02625 M
36. VPH2O° = 23.756 torr (25°C); m solute = 18.913 g; msolvent = 36 g H2O; VPsoln = 20.234 torr
MW solute = ?
, VPsoln = VPH2O°Xsolvent
20.234 = 23.756(x); x = 0.8517 therefore, the mole fraction of the solute is 0.1483
36 g/ 18.02g/mole = 1.9978 mole solvent; 0.8517 = 1.9978/(1.9978 + nsolute)
nsolute = 0.3479 moles
MW = 18.913 g/ 0.3479 moles = 54.3693 g/mol ~ 56 g/mol
37. 30% wt urea in water.
∆𝑇 = 𝐾𝑓𝑚𝑖
Since urea is non-electrolyte, i = 1;
m = (30 g urea/60.07 g/mol)/70 g H2O/1000g/kg
m = 7.1345 mol/kg
𝐶
∆𝑇 = (7.3145 𝑚) (1.86° ) (1) = 13.2702 𝐶°; Tsoln = 0-13.2702 = -13.2702°C
𝑚
38. m = 0.205 m with respect to urea
200 g of solution is diluted with 250 g of water
First find the number of moles of urea present in a 200 g solution.
0.205/1000 g = (x/60.07)/(200 – x); x = 2.4329 g of urea
m’ = (2.4329 g/60.07 g/mole)/(200 g – 2.4329 g + 250 g)/1000g/kg
m’ = 0.09049 m
𝐶
𝑇𝑓 = 0℃ − (0.09049 𝑚) (1.86° ) (1) = -0.1683 °C
𝑚
39. π = 38 mm Hg (at 273.15 K)
ΔTb =?
Π = MRT
M = (38mm Hg/760 mm Hg/1 atm)/(0.08206 atm-L/mol-K)(273.15 K)
M = 2.2307 x 10-3 M
∆𝑇𝑏 = 𝐾𝑏𝑚𝑖 𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑚 ~ 𝑀; = (0.512 𝐶° ) (2.2307 x 10−3 𝑚)(1) = 0.001142 𝐶°
𝑚
40. Ra-226 ----- He + 222 86Rn (α emission atomic mass decreased by 4, atomic number by 2)
41. Half-life is 29 years. Fraction (x) remained after 100 years = ?
Since half-life exhibits first order reactions,
k = ln2/t1/2 = ln2/29 = 0.0239 years-1
ln|Ao| - ln|A| = kt = ln|1| - ln|x| = 0.0239(100)
x = 0.0916 remains
42. Another first order reaction:
ln|1| - ln|0.01| = 50(k)
k = 0.092103 years-1
t1/2 = ln2/k = 7.5257 years
43. Ra-226 half-life = 1600 years
α emission; t= 10 mins; 10 mg sample of Ra-226. 1 year = 365 days.
Since it still follows first order reaction, k = ln2/1600 = 4.3322 x 10 -4 years-1
For t = 10 mins, 10/(60)(24)(365) = 1.9026 x 10-5 years thus,
kt = 8.24233 x 10-9 disintegration
for 10 mg, 10 mg/(226)(1000 mg/g) = 4.4248 x 10-5 moles = 2.6646 x 1019 particles
(2.6646 x 1019 particles)(8.24233 x 10-9 disintegration) = 2.1963 x 1011 disintegrations
44. Activity = number of disintegrations per second
1Ci = 3.7x1010 disintegration/atom-sec
k = ln2/(1600 x 365 x 24 x 3600) = 1.3737 x 10-11 sec-1
kn = (1.3737x10-11)(2.6646 x 1019) = 366041979.3 particles/s
Activity = 366041979.3 dps/3.7 x 1010 dps/Ci = 9.8930 Ci
FOR
CHEMICAL ENGINEERING
LICENSURE
EXAMINATION
3rd Edition
Solutions Manual
Monroe H. de Guzman, BSChE
Batangas State University
If some of the values are different, then these are brought by rounding off.
As much as possible, values are not rounded-off to obtain the most
possible accurate answers.
,
,I. Physical and Chemical Principles
A. General Inorganic Chemistry
4. For Zn2+, atomic number = 30, atomic mass = 65.38 (~65),
Number of protons = 30; neutrons = 65-30 = 35; electrons = 30 – 2 = 28
e- + proton + neutron = 30 + 35 + 28 = 93
6. The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six
different named series describing the spectral line emissions of the hydrogen atom (Wikipedia, 2014).
𝑛2
𝜆=( )(𝑅∞ /4); for m = 2 as n ∞
𝑛2−𝑚2
∞
Balmer series = 1/𝜆 = (𝑅∞ /4) = 10973731.57 m/4 = 27434.3289 cm -1
∞
7. Named after the American physicist Frederick Sumner Brackett who first observed the spectral
lines in 1922 (Wikipedia, 2014).
1 1
Brackett series = ( 2 − 2)(𝑅∞ ); m = 4 as n ∞ = 10973731.57 m/16 = 6858.5822 cm -1
𝑚 𝑛
10. Minimum wavelength of light for work functions (de Broglie wavelength)
For Φ = 2.90 eV (J/C of electron)
ℎ𝑐𝑜 (6.6261 𝑥 10−34 )(299792458)
𝜆= = = 427.5317 nm
𝜙𝑒 (2.90)(1.6022 𝑥 10−19 )
11. For Φ = 5.00 eV
ℎ𝑐𝑜 (6.6261 𝑥 10−34 )(299792458)
𝜆= = = 247.9584 nm
𝜙𝑒 (5.00)(1.6022 𝑥 10−19 )
13. Work function and kinetic energy
Rb Φ = 2.16 eV; λ = 350 nm of light wavelength.
Ephoton = Φ + Ek = hv = hco/λ; Ek = hco/λ – Φ
(6.6261 𝑥 10−34 )(299792458)
EK = − 2.16(1.6022 𝑥 10−19 )
350 𝑋 10−9
EK = 2.1667 x 10-19 J
Ek = (mev2)/2 = 689721.8665 m/s ~ 7.00 x 105 m/s
14. for n =2 and ms = -1/2. For l=0 and l = 1, it is composed of 3 counter-clockwise at p orbital and
1 at the s-orbital = 4 electrons
15. Cl = 35.45 composed of Cl-35 and Cl-37
35.45 = 37(x) + 35(1-x) where x is the fraction of Cl-37
x= 0.225; Cl-35 is 77.5 % in abundance.
17. If l = 0, ml should be = 0.
18. Outermost shell is at n = 4; taking the electrons at the outermost shell of 4s 2 and 4p3, 2 + 3=5.
31. T = 1600 °C; Br2(g) 2Br(g)
nBr2 = 1.05 moles ; Tv = 2 L; α = 0.025 Br2(g) 2Br(g)
[Br2] = 1.05/2 = 0.525 M
0.525 0
Kc = {([Br]2)/Br
2}eq = 1.3462 x M 10-3 0.525(1-α) 0.525x2α
Kp = Kc(RT)Σv
Kp = 1.3462 x 10-3(0.08206 x 1873.15)2-1 = 0.2069 atm 0.511875 M 0.02625 M
36. VPH2O° = 23.756 torr (25°C); m solute = 18.913 g; msolvent = 36 g H2O; VPsoln = 20.234 torr
MW solute = ?
, VPsoln = VPH2O°Xsolvent
20.234 = 23.756(x); x = 0.8517 therefore, the mole fraction of the solute is 0.1483
36 g/ 18.02g/mole = 1.9978 mole solvent; 0.8517 = 1.9978/(1.9978 + nsolute)
nsolute = 0.3479 moles
MW = 18.913 g/ 0.3479 moles = 54.3693 g/mol ~ 56 g/mol
37. 30% wt urea in water.
∆𝑇 = 𝐾𝑓𝑚𝑖
Since urea is non-electrolyte, i = 1;
m = (30 g urea/60.07 g/mol)/70 g H2O/1000g/kg
m = 7.1345 mol/kg
𝐶
∆𝑇 = (7.3145 𝑚) (1.86° ) (1) = 13.2702 𝐶°; Tsoln = 0-13.2702 = -13.2702°C
𝑚
38. m = 0.205 m with respect to urea
200 g of solution is diluted with 250 g of water
First find the number of moles of urea present in a 200 g solution.
0.205/1000 g = (x/60.07)/(200 – x); x = 2.4329 g of urea
m’ = (2.4329 g/60.07 g/mole)/(200 g – 2.4329 g + 250 g)/1000g/kg
m’ = 0.09049 m
𝐶
𝑇𝑓 = 0℃ − (0.09049 𝑚) (1.86° ) (1) = -0.1683 °C
𝑚
39. π = 38 mm Hg (at 273.15 K)
ΔTb =?
Π = MRT
M = (38mm Hg/760 mm Hg/1 atm)/(0.08206 atm-L/mol-K)(273.15 K)
M = 2.2307 x 10-3 M
∆𝑇𝑏 = 𝐾𝑏𝑚𝑖 𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑚 ~ 𝑀; = (0.512 𝐶° ) (2.2307 x 10−3 𝑚)(1) = 0.001142 𝐶°
𝑚
40. Ra-226 ----- He + 222 86Rn (α emission atomic mass decreased by 4, atomic number by 2)
41. Half-life is 29 years. Fraction (x) remained after 100 years = ?
Since half-life exhibits first order reactions,
k = ln2/t1/2 = ln2/29 = 0.0239 years-1
ln|Ao| - ln|A| = kt = ln|1| - ln|x| = 0.0239(100)
x = 0.0916 remains
42. Another first order reaction:
ln|1| - ln|0.01| = 50(k)
k = 0.092103 years-1
t1/2 = ln2/k = 7.5257 years
43. Ra-226 half-life = 1600 years
α emission; t= 10 mins; 10 mg sample of Ra-226. 1 year = 365 days.
Since it still follows first order reaction, k = ln2/1600 = 4.3322 x 10 -4 years-1
For t = 10 mins, 10/(60)(24)(365) = 1.9026 x 10-5 years thus,
kt = 8.24233 x 10-9 disintegration
for 10 mg, 10 mg/(226)(1000 mg/g) = 4.4248 x 10-5 moles = 2.6646 x 1019 particles
(2.6646 x 1019 particles)(8.24233 x 10-9 disintegration) = 2.1963 x 1011 disintegrations
44. Activity = number of disintegrations per second
1Ci = 3.7x1010 disintegration/atom-sec
k = ln2/(1600 x 365 x 24 x 3600) = 1.3737 x 10-11 sec-1
kn = (1.3737x10-11)(2.6646 x 1019) = 366041979.3 particles/s
Activity = 366041979.3 dps/3.7 x 1010 dps/Ci = 9.8930 Ci
